$a.3x^2+2x-1=0 \\⇔3x^2+3x-x-1=0 \\⇔3x(x+1)-(x+1)=0 \\⇔(x+1)(3x-1)=0$
⇔\(\left[ \begin{array}{l}x+1=0\\3x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-1\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=-1\\x=\dfrac{1}{3}\end{array} \right.\)
$b.x^2-5x+6=0 \\⇔x^2-2x-3x+6=0 \\⇔x(x-2)-3(x-2)=0 \\⇔(x-2)(x-3)=0$$⇔$\(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
$c.x^2-3x+2=0 \\⇔x^2-2x-x+2=0 \\⇔x(x-2)-(x-2)=0 \\⇔(x-2)(x-1)=0$$⇔$\(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
$d.2x^2-6x+1=0 \\⇔2\bigg(x^2-3x+\dfrac{1}{2}\bigg)=0 \\⇔x^2-3x+\dfrac{1}{2}=0 \\⇔x^2-2.\dfrac{3}{2}.x+\dfrac{9}{4}-\dfrac{7}{4}=0 \\⇔\bigg(x-\dfrac{3}{2}\bigg)^2-\dfrac{7}{4}=0 \\⇔\bigg(x-\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}\bigg)\bigg(x-\dfrac{3}{2}+\dfrac{\sqrt{7}}{2}\bigg)=0 \\⇔\bigg(x-\dfrac{3-\sqrt{7}}{2}\bigg)\bigg(x-\dfrac{3+\sqrt{7}}{2}\bigg)=0$
⇔\(\left[ \begin{array}{l}x-\dfrac{3-\sqrt{7}}{2}=0\\x-\dfrac{3+\sqrt{7}}{2}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{3-\sqrt{7}}{2}\\x=\dfrac{3+\sqrt{7}}{2}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=\dfrac{3-\sqrt{7}}{2}\\x=\dfrac{3+\sqrt{7}}{2}\end{array} \right.\)
